Solve 1/(x - 2) + 2/(x + 2) = 3/(x² - 4) for x.

Step-by-step solution

1. 1

   Factor the right-hand denominator and record the excluded values. Since $x^2-4=(x-2)(x+2)$, all three denominators are built from the same two factors, and the least common denominator is $(x-2)(x+2)$. The values $x=2$ and $x=-2$ make a denominator zero, so they can never be solutions — check any answer against them at the end.

2. 2

   Combine the two left-hand fractions. Rewrite each with the common denominator:

   $\frac{1}{x-2}+\frac{2}{x+2}=\frac{(x+2)+2(x-2)}{(x-2)(x+2)}$

3. 3

   Simplify the numerator.

   $(x+2)+2(x-2)=x+2+2x-4=3x-2$

   so the equation is now

   $\frac{3x-2}{(x-2)(x+2)}=\frac{3}{(x-2)(x+2)}$

4. 4

   Equate the numerators. The denominators are identical and non-zero for any admissible $x$, so the fractions are equal exactly when their numerators are:

   $3x-2=3\quad\Rightarrow\quad 3x=5\quad\Rightarrow\quad x=\frac{5}{3}$

5. 5

   Check against the excluded values and by substitution. $\frac53\neq 2$ and $\frac53\neq -2$, so the root is admissible. Substituting: the left side is $\frac{1}{5/3-2}+\frac{2}{5/3+2}=-3+\frac{6}{11}=-\frac{27}{11}$, and the right side is $\frac{3}{25/9-4}=\frac{3}{-11/9}=-\frac{27}{11}$. They agree, so $x=\frac53$.